Lecture on Rotational Motion in Classical Mechanics
Explore the concepts of rotational motion including rigid body motion, moment of inertia tensor, torque-free motion, comparison of analysis in inertial vs. non-inertial frames, and properties of frame motion (rotation) in this lecture. Understand the physics of rigid body motion, effects on acceleration, and kinetic energy of rigid bodies. Dive into the moment of inertia tensor, its matrix notation, and more.
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PHY 711 Classical Mechanics and Mathematical Methods 10-10:50 AM MWF Olin 103 Plan for Lecture 25: Rotational motion (Chapter 5) 1. Rigid body motion 2. Moment of inertia tensor 3. Torque free motion 10/24/2018 PHY 711 Fall 2018 -- Lecture 23 1
10/24/2018 PHY 711 Fall 2018 -- Lecture 23 2
10/24/2018 PHY 711 Fall 2018 -- Lecture 23 3
Comparison of analysis in inertial frame versus non- inertial frame 0 Denote by a fixed coordinate system Denote by a moving coordinate system i e e i 3 3 = = 0 0 V V For an arbitrary vector : V e Ve i i i i = = 1 1 i i 0 V dV dt dV dt de dt 3 3 3 d dt = = + 0 i e e i i i V i i = = = 1 1 1 i i i inertial V 3 dV dt d dt e Def ine: i i = 1 i body V V 3 de dt d dt d dt = + i V i = 1 i inertial body 10/24/2018 PHY 711 Fall 2018 -- Lecture 23 4
The physics of rigid body motion; body fixed frame vs inertial frame; results from Chapter 2: 10/24/2018 PHY 711 Fall 2018 -- Lecture 23 5
Properties of the frame motion (rotation): ze d = d e e ze d d y z = e e d d z y d = d e e d d dt e = e dt d y e d e y e d = e dt ( ) ( ( ) ) d e e e cos sin d d d y y y = ( ) e e e sin cos d d z z z 10/24/2018 PHY 711 Fall 2018 -- Lecture 23 6
V dt V dt e 3 d d d = i = + i V i dt 1 inertial body V dt V dt d d = + V inertial body Effects on acceleration: V dt V dt d d d d d = + + V dt dt inertial body body 2 2 V 2 V 2 V dt d d d dt = + + + V V 2 dt dt body inertial body 10/24/2018 PHY 711 Fall 2018 -- Lecture 23 7
Kinetic energy of rigid body : r r d d = + r r dt dt inertial body =0 for rigid body r d = r p dt inertial ( ) 2 1 1 p = = 2 p r T m v m p p p 2 2 ( )( ) 1 p = r r m p p p 2 )( ) ( ) 1 ( p 2 = r r r m p p p p 2 10/24/2018 PHY 711 Fall 2018 -- Lecture 23 8
1 2 )( ) ( ) m 2 ( = T r r r p p p p p 1 2 = I Moment I of inertia p 1 tensor r : ( ) 2 r (dyad notation) m p r p p p Matrix I notation : I I I xx xy xz r I I I yx yy yz I I I zx zy zz ( ) p 2 p I m r r ij p ij pi pj 10/24/2018 PHY 711 Fall 2018 -- Lecture 23 9
z Example: c y a b Moment of inertia + ab tensor a : x ( ) 2 2 b c ab ac 1 1 1 3 4 4 ( ) = + bc a 2 2 I M c bc 1 1 1 4 3 4 ( ) + 2 2 ac b 1 1 1 4 4 3 10/24/2018 PHY 711 Fall 2018 -- Lecture 23 10
Properties of moment of inertia tensor: Symmetric matrix real eigenvalues I1,I2,I3 orthogonal eigenvectors 3 , 2 , 1 = = i I i i i e e I Moment of inertia + ab tensor a : ( ) 2 2 b c ab ac 1 1 1 3 4 4 ( ) = + bc a 2 2 I M c bc 1 1 1 4 3 4 ( ) + 2 2 ac b 1 1 1 4 4 3 = = For : a b c 1 11 11 = = = 2 2 2 I Ma I Ma I Ma 1 2 3 6 12 12 10/24/2018 PHY 711 Fall 2018 -- Lecture 23 11
Changing origin of rotation z z ( ( ) p 2 p I m r r r ij p ij pi pj ) p 2 p ' ' ' ' I m r r r ij p ij pi pj rp = + r r R ' r p c p p R y Define center the m p of mass m p : y x r r a p p p p = b CM r m M p p x ( ) ( ) = + + 2 CM r R ' 2 I I M R R R M CMi r R R CMj r ij ij ij i j ij j i 10/24/2018 PHY 711 Fall 2018 -- Lecture 23 12
( ) ( ) = + + 2 CM r R ' 2 I I M R R R M CMi r R R CMj r ij z ij ij i j ij j i z = R x y z Suppose that a 2 b 2 c 2 = r ( R R and CM ) = 2 rp ' I I M R R r p ij ij ij i j c R y y ( ) x a + ab a 2 2 b c ab ac 1 1 1 3 4 4 ( ) b = + bc a 2 2 I ' M c bc 1 1 1 4 3 4 ( ) + 2 2 ac b 1 1 1 4 4 3 ( ) x + ab a 2 2 b c ab ac 1 1 1 4 4 4 ( ) + bc a 2 2 M c bc 1 1 1 4 4 4 ( ) + 2 2 ac b 1 1 1 4 4 4 10/24/2018 PHY 711 Fall 2018 -- Lecture 23 13
z Note: This is a special case; changing the center of rotation does not necessarily result in a diagonal I z rp r p c R y y x a b ( ) + 2 2 0 0 b c 1 x 12 ( ) = + 2 2 I ' 0 0 M a c 1 12 ( ) + 2 2 0 0 a b 1 12 10/24/2018 PHY 711 Fall 2018 -- Lecture 23 14
Descriptions of rotation about a given origin For general coordinate system 1 ij = T I ij i j 2 diagonaliz that system coordinate fixed) (body For es moment I of inertia tensor : = = e e e 3 , 2 , 1 I i i i i ~ ~ ~ 3 e = + + e e 1 1 1 2 2 3 3 2 e ~ i = 2 T I 1 e i i 2 10/24/2018 PHY 711 Fall 2018 -- Lecture 23 15
Descriptions of rotation about a given origin -- continued angular of change of rate Time L L dt dt body momentum d d = + L diagonaliz that system coordinate fixed) (body For es moment I I of inertia tensor = : + ~ ~ ~ = + e = e e e e I + I + 1 1 2 2 3 3 i i i ~ ~ ~ e e e L d I 1 1 1 2 2 2 3 3 3 L dt ( )3 e ) ~ ~ ~ ~ ~ = + + + e e e e I I I I I 1 ~ 1 ~ 1 2 2 2 3 ~ 3 ~ 3 2 3 3 2 1 ( ) ( + + e I I I I 3 1 1 3 2 1 2 2 1 10/24/2018 PHY 711 Fall 2018 -- Lecture 23 16
Descriptions of rotation about a given origin -- continued Note L dt that the = dt torque equation L d d + = L body very is difficult = solve to directly in the body fixed frame. For L dt we 0 can solve the Euler equations : d ( ) ) ~ ~ ~ ~ ~ = + + + e e e e I I I I I 1 1 1 2 2 2 3 3 3 2 3 3 2 1 ( ) ( ~ ~ ~ ~ + + = e e 0 I I I I 3 1 1 3 2 1 2 2 1 3 10/24/2018 PHY 711 Fall 2018 -- Lecture 23 17
Euler equations rotation for = I body in fixed frame : ( ( ( ) ) ) ~ ~ ~ + 0 I I 1 1 2 3 3 2 ~ ~ ~ + = 0 I I I 2 2 3 1 1 3 ~ ~ ~ + = 0 I I I 3 3 1 2 2 1 = Solution + symmetric for top - - : I I 2 1 ( ( ) ) ~ ~ ~ = 0 I I I 1 1 2 3 3 1 ~ ~ ~ + = 0 I I I 1 2 3 0 1 1 3 ~ ~ = = (constant) I I 3 3 3 ~ ~ = ~ I ~ 1 2 3 1 Define : ~ 3 = 1 I 1 2 10/24/2018 PHY 711 Fall 2018 -- Lecture 23 18
Solution of Euler equations for a symmetric top -- continued I I I t A t A t = 1 ~ 2 i ~ ~ e t A I + + = = = 1 2 2 1 where 3 1 3 1 = = + + Solution: 1 T = ( ) ( ) ( ) cos( sin( (constant) ~ 2 ~ e ) ) t 1 t 2 3 3 + 1 i = 2 i 2 2 3 I I A I 1 3 2 e = + + L e I I I 1 1 1 2 2 2 3 3 sin 3 ( ( ) ( ) ) ~ + + e e cos t I 1 1 2 3 3 3 10/24/2018 PHY 711 Fall 2018 -- Lecture 23 19
Euler equations rotation for = I body in fixed frame : ( ( ( ) ) ) ~ ~ ~ + 0 I I 1 1 2 3 3 2 ~ ~ ~ + = 0 I I I 2 2 3 1 1 3 ~ ~ Solution + ~ + = 0 I I I 3 3 1 2 1 asymmetric for 2 top - - : I I I 3 2 1 ( ( ( ) ) ) ~ ~ ~ = 0 I I I 1 1 2 3 3 2 ~ ~ ~ + = 0 I I I 2 2 3 1 1 3 ~ ~ ~ + = 0 I I I 3 3 1 2 2 1 I I ~ ~ 3 2 Suppose : 0 Define : 3 1 3 I 1 I I ~ 3 1 Define : 2 3 I 10/24/2018 PHY 711 Fall 2018 -- Lecture 23 20 2
Euler equations rotation for = I body in fixed frame : ( ( ( ) ) ) ~ ~ ~ + 0 I I 1 1 2 3 3 2 ~ ~ ~ + = 0 I I I 2 2 3 1 1 3 ~ ~ ~ + = 0 I I I 3 3 1 2 2 1 Solution for asymmetric top -- Approximate solution -- : I I I 3 2 1 I I Suppose: 0 Define: 3 2 3 1 3 I I 1 I Define: 3 1 2 3 I 2 10/24/2018 PHY 711 Fall 2018 -- Lecture 23 21
Euler equations for asymmetric top -- continued ( ( ( ) ) ) + = 0 I I I 1 1 2 3 3 2 + = 0 I I I 2 2 3 1 1 3 + = 0 I I I 3 3 1 2 2 1 I I I I If 0, Define: 3 2 3 1 3 1 3 2 3 I I 1 2 ~ ~ ~ ~ = = 1 1 2 2 2 1 If and A are both t positive or both negative : 1 2 ( ) ~ + ( ) cos t 1 1 2 ( ) ~ + ( ) sin 2 t A t 1 2 2 1 If and have opposite signs, solution unstable. is 1 2 10/24/2018 PHY 711 Fall 2018 -- Lecture 23 22
